\newproblem{lay:6_3_23}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.3.23}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A$ be an $m\times n$ matrix. Prove that every vector in $\mathbf{x}\in\mathbb{R}^n$ can be written in the form $\mathbf{x}=\mathbf{p}+\mathbf{u}$, where
	$\mathbf{p}$ is in $\mathrm{Row}\{A\}$ and $\mathbf{u}$ is in $\mathrm{Nul}\{A\}$. Also, show that if the equation $A\mathbf{x}=\mathbf{b}$ is consistent,
	then there is a unique $\mathbf{b}$ in $\mathrm{Row}\{A\}$ such that $A\mathbf{p}=\mathbf{b}$.
}{
   % Solution
	First, we'll show that $\mathrm{Row}\{A\}$ and $\mathrm{Nul}\{A\}$ are orthogonal subspaces. Let $\mathbf{a}_i$ ($i=1,2,...,m$) be the rows
	of matrix $A$. Any vector $\mathbf{u}$ in $\mathrm{Nul}\{A\}$ is such that
	\begin{center}
		$A\mathbf{u}=\mathbf{0}$
	\end{center}
	In particular, we may consider the multiplication of the $i$-th row of $A$ and $\mathbf{u}$
	\begin{center}
		$\mathbf{a}_i^T\mathbf{u}=0 \Rightarrow \mathbf{a}_i\cdot\mathbf{u}=0 \Rightarrow \mathbf{a}_i \perp \mathbf{u}$
	\end{center}
	
	Any vector $\mathbf{p}$ in $\mathrm{Row}\{A\}$ can be written as a linear combination of the rows of $A$
	\begin{center}
		$\mathbf{p}=c_1\mathbf{a}_1+c_2\mathbf{a}_2+...+c_m\mathbf{a}_m$
	\end{center}
	Let's calculate the inner product between $\mathbf{p}$ and $\mathbf{u}$
	\begin{center}
		$\mathbf{p}\cdot\mathbf{u}=c_1\mathbf{a}_1\cdot\mathbf{u}+c_2\mathbf{a}_2\cdot\mathbf{u}+...+c_m\mathbf{a}_m\cdot\mathbf{u}=0$
	\end{center}
	So $\mathbf{p}\perp\mathbf{u}$ for any $\mathbf{p}$ in $\mathrm{Row}\{A\}$ and any $\mathbf{u}$ in $\mathrm{Nul}\{A\}$.
	
	Since, both spaces are orthogonal to each other we may orthogonally project $\mathbf{x}\in\mathbb{R}^n$ onto $\mathrm{Row}\{A\}$ (obtaining $\mathbf{p}$) and
	onto $\mathrm{Nul}\{A\}$ (obtaining $\mathbf{u}$). By the Orthogonal Decomposition theorem we know that $\mathbf{x}$ can be uniquely decomposed as a vector in 
	$\mathrm{Row}\{A\}$ and a vector in $(\mathrm{Row}\{A\})^\perp=\mathrm{Nul}\{A\}$. This proves that $\mathbf{x}=\mathbf{p}+\mathbf{u}$.
	
	For the second part of the problem, let us presume that there are two distinct solutions $\mathbf{p}_1$ and $\mathbf{p}_2$ in $\mathrm{Row}\{A\}$ such that
	\begin{center}
		$A\mathbf{p}_1=\mathbf{b}$\\
		$A\mathbf{p}_2=\mathbf{b}$\\
	\end{center}
	Subtracting both equations we have
	\begin{center}
		$A(\mathbf{p}_1-\mathbf{p}_2)=\mathbf{0}$\\
	\end{center}
	That means that $\mathbf{p}_1-\mathbf{p}_2$ is in $\mathrm{Nul}\{A\}$. But at the same time it is in $\mathrm{Row}\{A\}$ (because it is the linear combination
	of two vectors in $\mathrm{Row}\{A\}$ and $\mathrm{Row}\{A\}$ is a vector space). The only vector that belongs both to $\mathrm{Nul}\{A\}$ and $\mathrm{Row}\{A\}$ is
	the zero vector so
	\begin{center}
		$\mathbf{p}_1-\mathbf{p}_2=\mathbf{0} \Rightarrow \mathbf{p}_1=\mathbf{p}_2$\\
	\end{center}
	which is a contradiction to our hypothesis that both solutions were distinct, and therefore, there is a single solution $\mathbf{p}$ in $\mathrm{Row}\{A\}$ of the
	problem $A\mathbf{p}=\mathbf{b}$.
}
\useproblem{lay:6_3_23}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
